Thread: Recursive and Special Sequence word problem

1. Recursive and Special Sequence word problem

Join two 1-unit by 1 unit squares to form a rectangle. Next, draw a larger square along a long side of the rectangle. Continue this process.

a) Write the sequence of the lengths of the sides of the squares you added at each step. Befin the sequence with two original squares.

b) Write a recursive formula for the sequence of lengths added.

c) Identify the sequence in b.

I dont know how to do any of it. step 1 has 2 squarws step 2 has 6 squares and step 3 has 15 square.

2. Re: Recursive and Special Sequence word problem

Join two 1-unit by 1 unit squares to form a rectangle. Next, draw a larger square along a long side of the rectangle. Continue this process.

a) Write the sequence of the lengths of the sides of the squares you added at each step. Befin the sequence with two original squares.

b) Write a recursive formula for the sequence of lengths added.

c) Identify the sequence in b.

I dont know how to do any of it. step 1 has 2 squarws step 2 has 6 squares and step 3 has 15 square.

Not quite sure as to what you mean.

1--Two 1 x 1 squares
2--Attach a 2 x 2 square to the 2 side of the rectangle making a 2 x 4 rectangle.
3--Attach a matching 4 x 4 square to the 4 side making a 4 x 8 rectangle.
4-- Attach a matching 8 x 8 square to the 8 side making an 8 x 16 rectangle.
5--...

1--1
2--8
3--32
4--128
5--...

1--2
2--4
3--8
4--16
5--...
????????????????????

Could you clarify?

3. Re: Recursive and Special Sequence word problem

Originally Posted by karliekay
Join two 1-unit by 1 unit squares to form a rectangle. Next, draw a larger square along a long side of the rectangle. Continue this process.

a) Write the sequence of the lengths of the sides of the squares you added at each step. Befin the sequence with two original squares.

b) Write a recursive formula for the sequence of lengths added.

c) Identify the sequence in b.

I dont know how to do any of it. step 1 has 2 squarws step 2 has 6 squares and step 3 has 15 square.
The number of squares increases by one each time, but I'm not using the number of squares for the pattern.

a) Sequence of the lengths of the sides:

1, 1, 2, 3, 5, 8, ...

b) $\ \ a_{n + 2} \ =\ a_{n + 1} \ + \ a_{n}, \ where \ a_0 \ = \ 1 \ and \ a_1 \ =\ 1.$

c) Fibonacci sequence

Did I show too much?

4. Re: Recursive and Special Sequence word problem

lookagain found the right path. Nice going.

Just as an afterthought, the nth term of the Fibonacci Sequence is given by
Fn = [((1 + sqrt5)/2)^n - ((1 -sqrt5)/2)^n]/sqrt(5)

If you are interested in some interesting tidbits about the Fibonacci Sequence, read on.

Closer examination of the series of numbers will lead you to some surprisingly interesting and unique relationships between the Fibonacci numbers Fn = F(n-1) + F(n-2).

As n increases, the ratio of F(n+1)/Fn approaches the Golden Ratio, (1 + sqrt5)/2 = 1.618....

The sum of the squares of two adjacent Fibonacci numbers is equal to a higher Fibonacci number according to Fn^2 + F(n+1)^2 = F(2n+1). For instance, the 4thFn^2 + the 5thFn^2 = the F(2(4) + 1) = 9th Fn or 3^2 + 5^2 = 34, the 9th Fn.

The product of two alternating Fibonacci numbers minus the square of the one in between is equal to +/- one as expressed by F(n-1)F(N+1) - Fn^2 = (-1)^n. For instance, 8x21 - 13^2 = 168 - 169 = (-1)^7 = -1.

The sum of the cubes of two adjacent Fibonacci numbers minus the cube of the preceding one is equal to a higher Fibonacci number as expressed by Fn^3 + F(n+1)^3 = F3n. For instance, F4^3 + F5^3 - F3^3 = 3^3 + 5^3 - 2^3 = 27 + 125 - 8 = 144 = F12 = 144.

The sum of the squares of a series of Fibonacci numbers starting with F1 and ending with Fn is equal to the product of Fn and F(n+1) as expressed by F1^2 + F2^2 + F3^2 + ......Fn^2 = FnF(n+1). For instance, 1^2 + 1^2 + 2^2 + 3^2 + 5^2 = 1 + 1 + 4 + 9 + 25 = 40 = 5x8.

The sum of a series of Fibonacci numbers starting with F1 and ending with Fn is equal to a higher Fibonacci number as expressed by F1 + F2 + F3 + .......Fn = F(n+2) - 1. For instance, 1 + 1 + 2 + 3 + 5 + 8 = 20 = 21 - 1.

The sum of any 10 consecutive Fibonacci numbers is 11 times the 7th term of the 10 numbers.
For instance, the sum of the 4th through 13th numbers, 3,5,8,13,21,34,55,89,144,233, is 11x55 = 605.

The sum of any number of consecutive Fibonacci numbers is given by S[Fn1-->Fn2] = F(n2+2) - F(n1+1).
For instance, the sum of the 5th through 10th numbers, 5,8,13,21,34,55, is 144 - 8 = 136.

The sum of the first n even terms of Fibonacci numbers is given by F2 + F4 + F6 + F8 + ....F2n = F(2n+1) - 1.
For instance, the sum of the first 6 even terms is therefore F(2(6) + 1) - 1 = F(13) - 1 = 233 - 1 = 232.

The sum of the first n odd terms of Fibonacci numbers is given by F1 + F3 + F5 + ....F(2n-1) = F(2n)
For instance, the sum of the first 6 odd terms is therefore F(2n) = F(12) = 144.

Fn^2 + F(n+1)^2 = F(2n+1)

F(n+1)F(n-1) - Fn^2 = (-1)^n

Some other relationships of Fibonacci numbers are Fn^2 + F(n+1)^2 = F(2n+1) and F(n+1)F(n-1) - Fn^2 = (-1)^n.

The ratio of the 2nth Fibonacci number divided by the nth Fibonacci number is always an integer or F2n/Fn = K. For instance, F10/F5 = 55/5 = 11.

The sum of any 4n consecutive Fibonacci numbers is evenly divisible by F2n.
Example: The sum of 4(2) = 8 Fibonacci numbers is divisible by F2(2) = F4 = 3.
1+2+3+5+8+13+21+34 = 87/3 = 29.
2+3+5+8+13+21+34+55 = 141/3 = 47.
3+5+8+13+21+34+55+89 = 228/3 = 76

As any number may be represented by combinations of powers of 2, so may any number be represented by combinations of the Fibonacci numbers.
1 = 1, 2 = 2, or 1+1, 3 = 3 or 1+2, 4 = 3+1, 5 = 5 or 3+2, 6 = 5+1 or 3+2+1, 7 = 5+2, 8 = 5+3, 9=5+3+1, 10 = 5+3+2, 50 = 34+13+3, 100 = 89+8+3 or 55+34+8+3, and so on.

The greatest common divisor of any two Fibonacci numbers is, itself, a Fibonacci number. Even more surprising is the fact that the g.c.d.(Fa, Fb) = c = F[g.c.d.(a,b)]. What this means is that the g.c.d. of the "a"th and "b"th Fibonacci numbers is the "c"th Fibonacci number where c = the g.c.d. of a and b.

For any two consecutive Fibonacci numbers, a and b, a^2 - ab - b^2 = +1 or -1.

Given the four consecutive Fibonacci numbers a, b, c and d, (ad), 2(bc) and (cd - ab) form a Pythagorean Triple such that (ad)^2 + [2(bc)]^2 = (cd - ab)^2.

Given two positive integers "m" and "n" that are relatively prime, the Fibonacci numbers F(m) and F(n) are also relatively prime.

5. Re: Recursive and Special Sequence word problem

Hello, karliekay!

Join two 1-by-1 squares to form a rectangle.
. . Next, draw a larger square along a long side of the rectangle.
. . Continue this process.

a) Write the sequence of the lengths of the sides of the squares you added at each step.
. . Begin the sequence with two original squares.

Code:
      *---*
1 |   |
*---*
1

Code:
      *---*
|   |
*---*
|:::| 1
*---*
1

Code:
      *---*-------*
1 |   |:::::::|
*---*:::::::| 2
1 |   |:::::::|
*---*-------*
2

Code:
      *---*-------*
1 |   |       |
*---*       | 2
1 |   |       |
*---*-------*
|:::::::::::|
|:::::::::::|
3 |:::::::::::| 3
|:::::::::::|
*-----------*
3

Code:
      *---*-------*-----------------*
1 |   |       |:::::::::::::::::|
*---*       |:::::::::::::::::|
1 |   |       |:::::::::::::::::|
*---*-------*:::::::::::::::::|
|           |:::::::::::::::::| 5
|           |:::::::::::::::::|
3 |           |:::::::::::::::::|
|           |:::::::::::::::::|
*-----------*-----------------*
3              5

$\text{It appears that the sequence is:}\;1,\,1,\,2,\,3,\,5,\,8,\,13\:\hdots$

b) Write a recursive formula for the sequence of lengths added.

$\text{Each term is the sum of the preceding two terms.}$

. . $F_n \;=\;F_{n-1} + F_{n-2}$

c) Identify the sequence in b.

$\text{This is the Fibonacci sequence.}$

6. Re: Recursive and Special Sequence word problem

That problem was my first exposure to Fibonacci - through old Scientific American - back in early 60's.

Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•